# Area of equilateral triangle given known distances of a point from the sides.

• October 28th 2009, 02:26 AM
findmehere.genius
Area of equilateral triangle given known distances of a point from the sides.
yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is
A
not uniquely determinable
B 6 root3
C 6
D 12 root3
• October 28th 2009, 05:24 AM
Wilmer
Label triangle A,B,C ; P = point inside
Place D on AB, E on BC, F on AC : PD = 2, PE = 3, PF = 1

Join PA, PB, PC

anglePAF = A ; then anglePAD = 60-A
sin(60-A) / sin(A) = 2

anglePBD = B ; then anglePBE = 60-B
sin(60-B) / sin(B) = 3/2

anglePCF = C ; then anglePCE = 60-C
sin(60-C) / sin(C) = 3

Plus if sides of triangle ABC = a, then area = 3a.

Hope that helps you continue on page 5 (Itwasntme)
• October 28th 2009, 05:46 AM
Hello findmehere.genius
Quote:

Originally Posted by findmehere.genius
yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is
A
not uniquely determinable
B 6 root3
C 6
D 12 root3

See the attached diagram.

Quadrilateral $ANXM$ is cyclic, since $\angle ANX = \angle AMX = 90^o$. So $\angle NMX = \angle NAX = \theta$, say. (Angles in same segment.)

Similarly $\angle NLX = \angle NBX = \phi$, say.

Now in $\triangle NMX, NM^2 = 2^2+3^3 - 2.2.3.\cos120^o = 19$ (Cosine Rule)

$\Rightarrow NM= \sqrt{19}$

So $\frac{\sin\theta}{2}=\frac{\sin120^o}{\sqrt{19}}$ (Sine Rule)

$\Rightarrow \sin\theta = \sqrt{\frac{3}{19}}$

$\Rightarrow \tan\theta = \frac{\sqrt3}{\sqrt{19-3}}=\frac{\sqrt3}{4}$

$\Rightarrow AN = \frac{2}{\tan\theta}$ (from $\triangle ANX$)

$= \frac{8}{\sqrt3}$

Similarly $AQ= \sqrt7$, $\sin\phi= \sqrt{\frac37}$, and $\tan\phi = \frac{\sqrt3}{2}$

$\Rightarrow NB = \frac{2}{\tan\phi}=\frac{4}{\sqrt3}$

$\Rightarrow AB = \frac{8}{\sqrt3}+\frac{4}{\sqrt3}=\frac{12}{\sqrt3 }=4\sqrt3$

So area $\triangle ABC = \tfrac12AB.AC\sin60^o=\tfrac12.4\sqrt3.4\sqrt3.\tf rac{\sqrt3}{2}=12\sqrt3$

• October 29th 2009, 07:53 AM
aidan
Quote:

Originally Posted by findmehere.genius
yet another question for someone willing to help for I am tired wasting four long pages....
Q.There is a point inside an equilateral triangle which is at distances
1, 2 and 3 from the three sides. The area of the triangle is

A
not uniquely determinable

B 6 root3

C 6

D 12 root3

In an equilateral triangle,
if you know the perpendicular distances to all three sides (which is given),
then that sum will equal the altitude (h) of the triangle (which is 1+2+3 = 6).

The base (or 1 side of the triangle) is $2 h \tan(30deg)$

$
\tan(30deg) = \dfrac{ \sqrt{3} } {3}
$

Thus:
$\text{Area} = h^2 \cdot \dfrac{ \sqrt{3}} {3}$ = $6^2 \cdot \dfrac{ \sqrt{3}} {3} \,\, = \,\, 12 \cdot \sqrt{3}$

as shown previously.

.